Question

Two conducting parallel plates are having charges $10\text{C}$ and $6\text{C}$ as shown in the figure. choose the correct statement(s) among the following.

• A. Charge on surface $\left(1\right)$ is $8\text{C}$.
• B. Charge on surface $\left(2\right)$ is $2\text{C}$.
• C. Charge on surface $\left(3\right)$ is $-2\text{C}$.
• D. Charge on surface $\left(4\right)$ is $8\text{C}$

Answer 1

Answer: (A), (B), (C), (D)

Charge on surface $\left(4\right)$ is $8\text{C}$

Given,
Net charge on plate $\text{A},Q_1=10\text{C}$
Net charge on plate $\text{B},Q_2=6\text{C}$

The charges $Q_1$ and $Q_2$ are distributed as shown below,


We know that, the charges present on surfaces $\left(1\right)$ and $\left(4\right)$ are equal. Similarly, charges present on surfaces $\left(2\right)$ and $\left(3\right)$ are equal and opposite.

Let $x$ be the charge present on surface $\left(3\right)$, so that $-x$ be the charge on surface $\left(2\right)$.

From the charges present on surfaces $\left(1\right)$ and $\left(4\right)$ We get, $Q_1+x=Q_2-x$

$\Rightarrow 2x=Q_2-Q_1$

$\Rightarrow x=\frac{Q_2-Q_1}{2}$

Substituting the values, we get

$x=\frac{6-10}{2}=-2\text{C}$

The charge present on surface $\left(3\right)$ is $q_3=-2\text{C}$

So, we can say that, the charge present on surface $\left(2\right)$ is $q_2=-x=+2\text{C}$

Charge present on surface $\left(1\right)$ is $q_1=Q_1+x=10-2=8\text{C}$

Charge present on surface $\left(4\right)$ is $q_1=Q_2-x=6-(-2)=8\text{C}$

Hence, options (a) , (b) , (c) and (d) are the correct answers.